Round off to 4 decimal places

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42.46% of 24.9653=10.60mol/cc approx.

So 10.6=24.9653(1-e⁻⁰˙⁰⁴ᵗ)+4.7938e⁻⁰˙⁰⁴ᵗ.

Let f(t)=24.9653(1-e⁻⁰˙⁰⁴ᵗ)+4.7938e⁻⁰˙⁰⁴ᵗ-10.6,

f(t)=24.9653-24.9653e⁻⁰˙⁰⁴ᵗ+4.7938e⁻⁰˙⁰⁴ᵗ-10.6,

f(t)=14.3650-20.1715e⁻⁰˙⁰⁴ᵗ

f'(t)=0.80686e⁻⁰˙⁰⁴ᵗ, f''(t)=-0.0322744e⁻⁰˙⁰⁴ᵗ.

Iteration formula applying method:

t=t-(14.3650-20.1715e⁻⁰˙⁰⁴ᵗ)/(0.80686e⁻⁰˙⁰⁴ᵗ),

t=t-(14.3650e⁰˙⁰⁴ᵗ-20.1715)/0.80686.

Use t₀=8, then:

t₁=8.48219,

t₂=8.4869,

t₃=8.4869.

So the solution is t=8.4869 (seconds?).

f(t₀)=-0.2825, f'(t₀)=0.5859, f''(t₀)=-0.02344 (all approx).

The second derivative is used in the convergence test. 

Let r be an exact root of f(t)=0, so t=r (f(r)=0) and:

Let t<n> mean the nth iteration of t.

We can write the iteration formula:

t<n+1>=t<n>-f(t<n>)/f'(t<n>).

Let ℇ<n> be the error in t<n>, so that r-t<n>=ℇ<n>.

We can also write:

r-t<n+1>=r-t<n>+f(t<n>)/f'(t<n>), that is:

ℇ<n+1>=ℇ<n>+f(t<n>)/f'(t<n>),

ℇ<n+1>=ℇ<n>+f(r-ℇ<n>)/f'(r-ℇ<n>).

Using Taylor series expansion:

ℇ<n+1>=ℇ<n>+(f(r)-ℇ<n>f'(r)+ℇ<n>²f''(r)/2+...)/(f'(r)-ℇ<n>f''(r)+...).

f(r)=0 by definition so:

ℇ<n+1>=ℇ<n>+(-ℇ<n>f'(r)+ℇ<n>²f''(r)/2+...)/(f'(r)-ℇ<n>f''(r)+...),

ℇ<n+1>≈ℇ<n>+(-ℇ<n>f'(r)+ℇ<n>²f''(r)/2)/(f'(r)(1-ℇ<n>f''(r)/f'(r)),

ℇ<n+1>≈ℇ<n>+(-ℇ<n>+ℇ<n>²f''(r)/(2f'(r))/(1-ℇ<n>f''(r)/f'(r)),

ℇ<n+1>≈ℇ<n>+(-ℇ<n>+ℇ<n>²f''(r)/(2f'(r))(1+ℇ<n>f''(r)/f'(r)),

ℇ<n+1>≈ℇ<n>-ℇ<n>+ℇ<n>²f''(r)/(2f'(r))-ℇ<n>²f''(r)/f'(r)-ℇ<n>³f''(r)).

We can ignore the cube and subsequent terms as being insignificant, so:

ℇ<n+1>≈-ℇ<n>²f''(r)/(2f'(r)).

We can see that f''(t)/(2f'(t))=-0.0322744/1.61372=-1/50=-0.02. This may be the definition of g(t), which is constant, independent of t.

So ℇ<n+1>ℇ<n>²/50, so the error is rapidly decreasing with each iteration. This proves convergence.

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